Given equation is \(y=2\cos { 2\pi } \left( 10t-0.0080x+3.5 \right) \)  
Comparing with standard equation,
\(y=2\cos { 2\pi } \left( 10t-0.0080x+3.5 \right) \)
\(y=a\cos { \left( \omega t-kx+\phi \right) } \)
\( a=2cm,\omega =\frac { 2\pi }{ T } =20\pi ,\ T=0.1s\)
\( k=\frac { 2\pi }{ \lambda } =0.008\times 2\pi \Rightarrow \lambda =\frac { 2\pi }{ 2\pi \times 0.008 } =1.25m\)
\(\\ \phi =2\pi \times 3.5=7\pi rad\)
\(At\ t=T;\phi =\frac { 2\pi }{ T } =\frac { 2\pi }{ 0.1 } =20\pi rad\) 
\(\\ and\ at\ t=5s;{ \phi }^{ ' }=\frac { 2\pi }{ 0.1 } \times 5.100\pi rad\)
\(\therefore phase\ difference\ { \phi }^{ ' }-\ \phi =100\pi rad-20\pi rad=8020\pi rad\).

\(\omega =\frac { 2\pi }{ T } =100\pi \)
\( K=\frac { 2\pi }{ \lambda } =0.4\pi \Rightarrow \lambda =\frac { 2 }{ 0.4 } =\frac { 20 }{ 4 } =5m\)

A: Find amplitude of the given wave.
Given, progressive wave,
y=5sin(100πt−0.4πx)
So, amplitude of the given wave,
a=5 m
Find wavelength of the given wave.
Since we know
k=2π/λ
According to the given wave equation,
Angular wavenumber, k=0.4π
2π/λ=0.4π
λ=20/4=5 m
Final answer: 5m;5m
B: Find frequency of the wave
Given, wave equation,
y=5sin(100πt−0.4πx)
From the given waveequation, angular frequency,
ω=100π
Since ω=2πf
So,frequency,
f=ω²/π
\(\omega =2\pi v=100\pi v=50Hz\)
f=100π/2π=50 Hz
Find wave velocity of the wave.
Given, wavelength of the wave,
λ=5m
Wave velocity, v=fλ
v=50×5=250m/s
Final Answer: 50 Hz;250m/s
C: Particle velocity: Particle velocity is the velocity with which particles vibrate during the wave propagation.
Given, wave equation,
y=5sin(100πt−0.4πx)
Velocity of the particle by, \(\frac{\partial_{\mathrm{y}}}{\partial_{\mathrm{t}}}=\frac{\partial}{\partial_{\mathrm{t}}}(5 \sin (100 \pi \mathrm{t}-0.4 \pi \mathrm{t})) \)
\( \frac{\partial_y}{\partial_t}=500 \pi \cos (100 \pi t-0.4 \pi x)\)
So, particle velocity amplitude = \(500 \pi \mathrm{m} / \mathrm{s}\)
Final Answer: \(500 \pi \mathrm{m} / \mathrm{s}\)